Closedness of the product of two subgroups Let $G:=\text{SL}(3,\mathbb R)$, equipped with the usual subspace topology, acting on $\mathbb R^3$ by the canonical action. Consider subgroups $\Gamma:=\text{SL}(3,\mathbb Z)$ and $Q_1:=\{g\in G:ge_1=e_1\}$ where $e_1=(1,0,0)$. I wonder how to prove that the product $\Gamma Q_1$ is a closed subset of $G$.
To begin with, let $\gamma_n q_n$ be a sequence in $\Gamma Q_1$ ($\gamma_n \in \Gamma$ and $q_n\in Q_1$) converging to $h\in G$. Consider $\gamma_n q_n e_1=\gamma_n e_1$. We can conclude from here that the first column of $\gamma_n$ with stabilize at some point. But I don't know how to proceed from here. Any other approaches will also be appreciated! 
 A: Let $\pi$ be the projection $G\to G/Q_1$. 
Then $\Gamma Q_1$ is closed in $G$ iff the orbit $\Gamma\pi(1)$ is closed in $G/Q_1$. Now observe that the map $g\mapsto ge_1$ induces an identification of $G/Q_1$ with $\mathbf{R}^3\smallsetminus\{0\}$, and in this identification, $\pi(1)=e_1$. Now we see that the orbit of $e_1$ is the set of primitive elements in $\mathbf{Z}^3$, which is obviously closed. This proves the result.

Now this can be translated to a more pedestrian proof.
Let $g_n=\gamma_nq_n$ converge to $h$. Then $\gamma_ne_1=\gamma_nq_ne_1$ tends to $he_1$. Since $\gamma_ne_1$ converges and also belongs to the closed discrete subset $\mathbf{Z}^3$, we deduce that for $n$ large enough (as we now assume), it is constant, say equal to $\gamma e_1$ for some given $\gamma$. So $\gamma_ne_1=\gamma e_1$, which implies $\gamma^{-1}\gamma_n\in Q_1$, that is, $\gamma_n=\gamma q'_n$. So $\gamma^{-1}g_n=q'_nq_n$ converges, say to $q\in Q_1$. Thus $g_n$ converges to $\gamma q$, and hence $h=\gamma q\in \Gamma Q_1$.
